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3 years ago

15

What is an equation of the line that passes through the points : (−6,−5) and (−6,−2)?

1 answer:

Leno4ka [110]3 years ago

5 0

Answer: x= -6

Step-by-step explanation:

It will be a vertical line parallel to the y-axis at -6 on the x-axis. It passes through every y-value, including the given -5 and -2

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What are the solutions to the equation 3(x – 4)(x + 5) = 0?

iris [78.8K]

Answer:

x=4 and x= - 5

Step-by-step explanation:

In order to solve this we should re-arrange the equation:

$(x-4)(x+5)=\frac{0}{3}$

This is equal to:

$(x-4)(x+5)=0$

Then we can seperate this into:

$(x-4)=0\\(x+5)=0$

So solving for x in both cases we get

$x-4=0 \\x=4\\And\\x+5=0\\x=-5\\$

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2 years ago

One-third of a number x is equal to 22 less than the number

algol [13]

One-third of a number x is equal to 22 less than the number.

(1/3)x=x-22

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3 years ago

What number complete the square x^2 +\- bx

kati45 [8]

Take 1/2 of b and squaer it so (b/2)^2 would complete the square (sign doesn't matter becaus when squareing, sign goes positive)

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3 years ago

The endpoints of a diameter of a circle are (6,2) and (−2,5) . What is the standard form of the equation of this circle?

Paul [167]

Answer:

The standard form of the given circle is

$(x-2)^2+(y-\frac{7}{2})^2=73$

Step-by-step explanation:

Given that the end points of a diameter of a circle are (6,2) and (-2,5);

Now to find the standard form of the equation of this circle:

The center is (h,k) of the circle is the midpoint of the given diameter

midpoint formula is $M=(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$

Let $(x_{1},y_{1})$ and $(x_{2},y_{2})$ be the given points (6,2) and (-2,5) respectively.

$M=(\frac{6-2}{2},\frac{2+5}{2})$

$M=(\frac{4}{2},\frac{7}{2})$

$M=(2,\frac{7}{2})$

Therefore the center (h,k) is $(2,\frac{7}{2})$

now to find the radius:

The diameter is the distance between the given points (6,2) and (-2,5)

$d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

$=\sqrt{(-2-6)^2+(5-2)^2}$

$=\sqrt{(-8)^2+(3)^2}$

$=\sqrt{64+9}$

$=\sqrt{73}$

Therefore the radius is $\sqrt{73}$

i.e., $r=\sqrt{73}$

Therefore the standard form of the circle is

$(x-h)^2+(y-k)^2=r^2$

Now substituting the center and radiuswe get

$(x-2)^2+(y-\frac{7}{2})^2=(\sqrt{73})^2$

$(x-2)^2+(y-\frac{7}{2})^2=73$

Therefore the standard form of the given circle is

$(x-2)^2+(y-\frac{7}{2})^2=73$

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3 years ago

The range of the following relation R {(3, -5), (1,2),(-1,-4),(-12)}

lukranit [14]

Answer:

The range of the following relation is(-5,2,-4)

6 0

2 years ago